What is the limit as h approaches 0 of ((1+h)^5-1)/h ?

The limit as h approaches 0 of ((1+h)^5-1)/h is 5

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limit as h approaches 0 of ((1+h)^5-1)/h

h \to 0 lim (\frac{( 1 + h ) ^{5} - 1}{h})

5

Solution steps

h \to 0 lim (\frac{( 1 + h ) ^{5} - 1}{h})

Simplify

\frac{( 1 + h ) ^{5} - 1}{h} : h + (h + 1)^{4} + (h + 1)^{3} + (h + 1)^{2} + 2

= h \to 0 lim (h + (h + 1)^{4} + (h + 1)^{3} + (h + 1)^{2} + 2)

Plug in the value

h = 0

= 0 + (0 + 1)^{4} + (0 + 1)^{3} + (0 + 1)^{2} + 2

Simplify

0 + (0 + 1)^{4} + (0 + 1)^{3} + (0 + 1)^{2} + 2 : 5

= 5

\int \frac{t ^{2}}{t} d t

(x^{2} + 1) y^{^{'}} + 5 x (y - 1) = 0

\int_{- 1}^{3} x + 1 d x

\frac{\partial}{\partial y} (2 x^{3} y + 3)

\int_{- 3}^{- 5} - 4^{- 3} d x

What is the limit as h approaches 0 of ((1+h)^5-1)/h ?
The limit as h approaches 0 of ((1+h)^5-1)/h is 5